Example 1: Find the inverse function, its domain and range, of the function given by
f(x) = √(x  1)
Solution to example 1:

Note that the given function is a square root function with domain [1 , + ∞) and range [0, +∞). We first write the given function as an equation as follows
y = √(x  1)

Square both sides of the above equation and simplify
y^{ 2} = (√(x  1))^{ 2}
y^{ 2} = x  1

Solve for x
x = y^{ 2} + 1

Change x into y and y into x to obtain the inverse function.
f^{ 1}(x) = y = x^{ 2} + 1
The domain and range of the inverse function are respectively the range and domain of the given function f. Hence
domain and range of f^{ 1} are given by: domain: [0,+ ∞) range: [1 , + ∞)
Example 2: Find the inverse, its domain and range, of the function given by
f(x) = √(x + 3)  5
Solution to example 2:

Let us first find the domain and range of the given function.
Domain of f: (x + 3) ≥ 0 which gives x ≥  3 and in interval form [ 3 , + ∞)
Range of f: [ 5 , +∞)

Write f as an equation.
y = √(x + 3)  5
which gives √(x + 3) = 5 + y

Square both sides of the above equation and simplify.
(√(x + 3))^{ 2} = (5 + y)^{ 2}
(x + 3) = (5 + y)^{ 2}

Solve for x.
x = (5 + y)^{ 2}  3

Interchange x and y to obtain the inverse function
f^{ 1}(x) = y = (5 + x)^{ 2}  3
The domain and range of the inverse function are respectively the range and domain of the given function f. Hence
domain and range of f^{ 1} are given by: domain: [ 5,+ ∞) range: [ 3 , + ∞)
Example 3: Find the inverse, its domain and range, of the function given by
f(x) =  √(x^{ 2} 1) ; x ≤ 1
Solution to example 3:

Function f given by the formula above is an even function and therefore not a one to one if the domain is the set R. However the domain in our case is given by x ≤  1 which makes the given function a one to one and therefore has inverse.
Domain of f: ( ∞ ,  1] , given
Range: For x in the domain ( ∞ ,  1] , the range of x^{ 2}  1 is given by [0,+∞), which gives a range of f(x) =  √(x^{ 2} 1) in the interval ( ∞ , 0].

Write f as an equation, square both sides and solve for x, and find the inverse.
y =  √(x^{ 2} 1)
y^{ 2} = ( √(x^{ 2} 1))^{ 2}
y^{ 2} = x^{ 2} 1
x^{ 2} = y^{ 2} + 1
x = ~+mn~√(y^{ 2} + 1)

We now apply the domain of f given by x ≤ 1 to select one of the two solutions above. Hence
x =  √(y^{ 2} + 1)

Change x into y and y into x to obtain the inverse function.
f^{1}(x) = y =  √(x^{ 2} + 1)
The domain and range of f^{ 1} are respectively given by the range and domain of f found above
Domain of f^{ 1} is given by: [0 , + ∞) and its range is given by: ( ∞ , 1]
Exercises: Find the inverse, its domain and range, of the functions given below
1. f(x) = 2 √(x + 2)  6
2. g(x) = 2 √(x^{ 2}  4) + 4 ; x ≥ 2
Answers to above exercises:
1. f^{ 1}(x) = (1/4)(x + 6)^{ 2}  2 ; domain: (∞ ,  6] Range: [ 2 ; ∞)
2. g^{ 1}(x) = √((y  4)^{ 2} / 4 + 4) ; domain: [4 , +∞) Range: [2 , +∞)
More links and references related to the inverse functions.
Find the Inverse of a Square Root Function
Find the Inverse Functions  Calculator
Applications and Use of the Inverse Functions
Find the Inverse Function  Questions
Find the Inverse Function (1)  Tutorial.
Definition of the Inverse Function  Interactive Tutorial
Find Inverse Of Cube Root Functions.
Find Inverse Of Square Root Functions.
Find Inverse Of Logarithmic Functions.
Find Inverse Of Exponential Functions.
